Problem: Solve for $q$, $ -\dfrac{6}{25q - 20} = \dfrac{3q + 2}{10q - 8} + \dfrac{8}{25q - 20} $
First we need to find a common denominator for all the expressions. This means finding the least common multiple of $25q - 20$ $10q - 8$ and $25q - 20$ The common denominator is $50q - 40$ To get $50q - 40$ in the denominator of the first term, multiply it by $\frac{2}{2}$ $ -\dfrac{6}{25q - 20} \times \dfrac{2}{2} = -\dfrac{12}{50q - 40} $ To get $50q - 40$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ \dfrac{3q + 2}{10q - 8} \times \dfrac{5}{5} = \dfrac{15q + 10}{50q - 40} $ To get $50q - 40$ in the denominator of the third term, multiply it by $\frac{2}{2}$ $ \dfrac{8}{25q - 20} \times \dfrac{2}{2} = \dfrac{16}{50q - 40} $ This give us: $ -\dfrac{12}{50q - 40} = \dfrac{15q + 10}{50q - 40} + \dfrac{16}{50q - 40} $ If we multiply both sides of the equation by $50q - 40$ , we get: $ -12 = 15q + 10 + 16$ $ -12 = 15q + 26$ $ -38 = 15q $ $ q = -\dfrac{38}{15}$